Problem: Find $ \lim_{x\to -3}g(x)$ for $g(x)=\sqrt{7x+22}$.
$g$ is a square-root function. Square-root functions are continuous across their entire domain, and their domain is all real $x$ -values for which the expression within the square-root is non-negative. In other words, for any square-root function $q$ and any input $c$ in the domain of $q$ (except for its endpoint), we know that this equality holds: $\lim_{x\to c}q(x)=q(c)$ [What happens at the endpoint?] The input $x=-3$ is within the domain of $g$. Therefore, in order to find $ \lim_{x\to -3}g(x)$, we can simply evaluate $g$ at $x=-3$. $\begin{aligned} &\phantom{=}g(x) \\\\ &=\sqrt{7x+22} \\\\ &=\sqrt{7(-3)+22} \gray{\text{Substitute }x=-3} \\\\ &=\sqrt{1} \\\\ &=1 \end{aligned}$ In conclusion, $ \lim_{x\to -3}g(x)=1$.